Question

我有四个音频。让我们说s1，s2，s3和s4。我还有一个变量，其中一个和四个之间的随机整数，例如myVar = [2,1,3,4]。它可以长于或短于四个，但其中的整数总是在1到4之间。

我想创建一个循环，它将在myVar上查看并以相应的顺序播放我的声音（在这种情况下：s2，s1，s3，s4）。一个接一个的声音，这是非常重要的。不是一次全部。

到目前为止，我的所有解决方案都会立即播放声音。我知道有几个类似的问题，但我在JavaScript上非常糟糕，我无法使用答案。有谁知道怎么做？

当前解决方案之一（不起作用）：

library(dplyr)

dft1 <- dft1 %>%
        mutate(Presence = case_when(# If R.Detect is "moved" or "open" or "closed"
                                    # AND (   (EP.Detect on previous row is "Movement" AND LP.Detect on previous row is "Movement" or "TimeOut)
                                    #      OR (EP.Detect on previous row is "TimeOut" AND LP.Detect on previous row is "Movement")
                                    #     )
                                    # THEN set Presence to 1
                                    .$R.Detect %in% c("moved","open", "closed")
                                    & (  (lag(.$EP.Detect) == "Movement" & lag(.$LP.Detect) %in% c("Movement","TimeOut"))
                                       | (lag(.$EP.Detect) == "TimeOut" & lag(.$LP.Detect) == "Movement")
                                      ) ~ 1, 
                                    # All other cases, set Presence to 0
                                    TRUE ~ 0 
                                   )
              )

Answer 1

使用onended事件播放下一首歌，如下所示：

var musics = ['s1', 's2', 's3', 's4'];
var order = [2,1,3,4];
var player; //get your audio tag here
player.autoplay = true;

function playNext() {
    var current = order.shift();
    player.src = musics[current];
}

player.addEventListener('ended', playNext);
playNext();

Answer 2

也可以有一个没有'order'数组突变的解决方案，例如使用Symbol.iterator为每个ended事件调用：

var a1 = new Audio('http://developer.mozilla.org/@api/deki/files/2926/=AudioTest_(1).ogg'); a1.id = "1"; // id helps to see console.log is fine
var a2 = new Audio('http://developer.mozilla.org/@api/deki/files/2926/=AudioTest_(1).ogg'); a2.id = "2";
var a3 = new Audio('http://developer.mozilla.org/@api/deki/files/2926/=AudioTest_(1).ogg'); a3.id = "3";
var a4 = new Audio('http://developer.mozilla.org/@api/deki/files/2926/=AudioTest_(1).ogg'); a4.id = "4";

var order = [2, 1, 4, 3, 2, 1];

rotateAudio(order); // main function; pass order array here

function rotateAudio(order) {
  var orderMap = {1: a1, 2: a2, 3: a3, 4: a4};
  var iterator = order[Symbol.iterator]();
  var value, done;
  
  function rotate({value, done}) { // get next iteration
    function next(e) {
      e.target.removeEventListener('ended', next); // helps for garbage removal I guess
      rotate(iterator.next())
    };
    if (done) { console.log('done!'); return };
    var current = orderMap[value];
    console.log('current: ', current);
    current.play();
    current.addEventListener('ended', next) // listen to 'ended' on audio, start next one
  }
  
  rotate(iterator.next()); // first call
}

Answer 3

使用.onended代替.addEventListener，它非常可靠。

var musics = ['s1', 's2', 's3', 's4'];
var playArray=function(){
  if(order[]!=null){
    var voice =musics[0];
     voice.play();
    order.shift();
    voice.onended = function () {
         playArray();
    };
  }
}

循环播放一个接一个的声音

3 个答案: